" The kernel is important because it controls the entire homomorphism. It tells us not only which elements of G are mapped to the identity in G', but also which pairs of elements have the same image in G'. " (Taken from Algebra by Artin.)

How does the kernel tell us which pairs of elements have the same image in G'?

What I tried:

let $f$ be a homomorphism such that $f(a) = f(b) = c$. $c$ is in $G'$, so it has an inverse $c^{-1}$. So then, $c^{-1}f(a) = c^{-1}f(b) = cc^{-1} = 1$.

But, this is tedious to compute, and I didn't use the kernel to do it.

  • 2
    $\begingroup$ $f(a)=f(b)=c\implies f(ab^{-1})=e'$. $\endgroup$ – lulu Jun 9 '19 at 18:51

The point is that $f(a)=f(b)$ if and only if $ab^{-1}\in \ker f$. So, if you have checked beforehand that $\ker f=\{e\}$, then $f(a)=f(b)$ automatically implies that $a=b$, which means that $f$ is injective. And it works backwards too, if $f$ is injective, then $\ker f=\{e\}$, because $f(a)=e =f(e)$ implies $a=e$.


If $c$ belongs to the image, then there is a $g_0\in G$ such that $f(g_0)=c$. But then$$\{g\in G\mid f(g)=c\}=\{g_0h\mid h\in\ker f\}.$$

  • $\begingroup$ does this imply that $g = g_0h$? (for every $h \; \epsilon $ ker $f$ ?) $\endgroup$ – Jess Jun 9 '19 at 18:58
  • $\begingroup$ What is $g$? What is $h$? $\endgroup$ – José Carlos Santos Jun 9 '19 at 18:59
  • $\begingroup$ $g$ is every element of $G$ such that $f(g) = c$. $h$ is any element of $G$ in the Kernel? $\endgroup$ – Jess Jun 9 '19 at 19:01
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    $\begingroup$ What I am saying is that an element $g$ of $G$ has the property that $f(g)=c$ when and only when $g$ can be written as $g_0h$, where $h$ is an element of $\ker f$. $\endgroup$ – José Carlos Santos Jun 9 '19 at 19:03

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